1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Find the roots of each of the following equations, if they exist, by applying the quadratic formula: ${a}^{2}{b}^{2}{x}^{2}-\left(4{b}^{4}-3{a}^{4}\right)x-12{a}^{2}{b}^{2}=0$, $a\ne 0$ and $b\ne 0$ [CBSE 2006]

Open in App
Solution

## The given equation is ${a}^{2}{b}^{2}{x}^{2}-\left(4{b}^{4}-3{a}^{4}\right)x-12{a}^{2}{b}^{2}=0$. Comparing it with $A{x}^{2}+Bx+C=0$, we get A = ${a}^{2}{b}^{2}$, B = $-\left(4{b}^{4}-3{a}^{4}\right)$ and C = $-12{a}^{2}{b}^{2}$ ∴ Discriminant, D = ${B}^{2}-4AC={\left[-\left(4{b}^{4}-3{a}^{4}\right)\right]}^{2}-4×{a}^{2}{b}^{2}×\left(-12{a}^{2}{b}^{2}\right)=16{b}^{8}-24{a}^{4}{b}^{4}+9{a}^{8}+48{a}^{4}{b}^{4}=16{b}^{8}+24{a}^{4}{b}^{4}+9{a}^{8}={\left(4{b}^{4}+3{a}^{4}\right)}^{2}>0$ So, the given equation has real roots. Now, $\sqrt{D}=\sqrt{{\left(4{b}^{4}+3{a}^{4}\right)}^{2}}=4{b}^{4}+3{a}^{4}$ $\therefore \alpha =\frac{-B+\sqrt{D}}{2A}=\frac{-\left[-\left(4{b}^{4}-3{a}^{4}\right)\right]+\left(4{b}^{4}+3{a}^{4}\right)}{2×{a}^{2}{b}^{2}}=\frac{8{b}^{4}}{2{a}^{2}{b}^{2}}=\frac{4{b}^{2}}{{a}^{2}}\phantom{\rule{0ex}{0ex}}\beta =\frac{-B-\sqrt{D}}{2A}==\frac{-\left[-\left(4{b}^{4}-3{a}^{4}\right)\right]-\left(4{b}^{4}+3{a}^{4}\right)}{2×{a}^{2}{b}^{2}}=\frac{-6{a}^{4}}{2{a}^{2}{b}^{2}}=-\frac{3{a}^{2}}{{b}^{2}}$ Hence, $\frac{4{b}^{2}}{{a}^{2}}$ and $-\frac{3{a}^{2}}{{b}^{2}}$ are the roots of the given equation.

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Nature of Roots
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program